Comparing Fractions - ISEE Lower Level Quantitative Reasoning

A painter has two cans of paint of the same size. The can of red paint is $\frac{5}{12}$ full. The can of blue paint is $\frac{4}{9}$ full. Which statement accurately compares the amounts of paint?
Answer: There is more blue paint than red paint. To compare the fractions $\frac{5}{12}$ and $\frac{4}{9}$ , find a common denominator. The least common multiple of 12 and 9 is 36. Convert each fraction to have a denominator of 36. For the red paint: $\frac{5}{12} = \frac{5 \times 3}{12 \times 3} = \frac{15}{36}$ . For the blue paint: $\frac{4}{9} = \frac{4 \times 4}{9 \times 4} = \frac{16}{36}$ . Since $16 > 15$ , $\frac{16}{36} > \frac{15}{36}$ . Thus, there is more blue paint than red paint.
A water tank was $\frac{4}{5}$ full. After a day of use, it was $\frac{1}{3}$ full. A different, identical tank was $\frac{5}{6}$ full and after a day was $\frac{1}{2}$ full. Which tank had a greater fraction of its water used?
Answer: The first tank First, calculate the fraction of water used from each tank. For the first tank, the amount used is $\frac{4}{5} - \frac{1}{3}$ . The common denominator is 15: $\frac{12}{15} - \frac{5}{15} = \frac{7}{15}$ . For the second tank, the amount used is $\frac{5}{6} - \frac{1}{2}$ . The common denominator is 6: $\frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3}$ . Now, compare the fractions of water used: $\frac{7}{15}$ and $\frac{1}{3}$ . The common denominator is 15: $\frac{1}{3} = \frac{5}{15}$ . Since $\frac{7}{15} > \frac{5}{15}$ , the first tank had a greater fraction of its water used.
Three of the following fractions are greater than $\frac{2}{3}$ . Which fraction is NOT greater than $\frac{2}{3}$ ?
Answer: $\frac{8}{13}$ We need to compare each fraction to $\frac{2}{3}$ to find the one that is smaller. We can use cross-multiplication. A) For $\frac{3}{4}$ , $3 \times 3 = 9$ and $4 \times 2 = 8$ . Since $9 > 8$ , $\frac{3}{4} > \frac{2}{3}$ . B) For $\frac{5}{7}$ , $5 \times 3 = 15$ and $7 \times 2 = 14$ . Since $15 > 14$ , $\frac{5}{7} > \frac{2}{3}$ . C) For $\frac{7}{10}$ , $7 \times 3 = 21$ and $10 \times 2 = 20$ . Since $21 > 20$ , $\frac{7}{10} > \frac{2}{3}$ . D) For $\frac{8}{13}$ , $8 \times 3 = 24$ and $13 \times 2 = 26$ . Since $24 < 26$ , $\frac{8}{13} < \frac{2}{3}$ . Thus, $\frac{8}{13}$ is the fraction that is not greater than $\frac{2}{3}$ .
Three ropes have lengths of $\frac{5}{9}$ meter, $\frac{1}{2}$ meter, and $\frac{4}{7}$ meter. Which rope's length is between the lengths of the other two?
Answer: The $\frac{5}{9}$ meter rope To find the rope with the middle length, we must order the fractions $\frac{5}{9}$ , $\frac{1}{2}$ , and $\frac{4}{7}$ . One way is to convert them to decimals: $\frac{1}{2} = 0.5$ , $\frac{5}{9} = 0.555...$ , and $\frac{4}{7} \approx 0.571$ . Ordering these from least to greatest gives 0.5, 0.555..., 0.571. This corresponds to the order $\frac{1}{2}$ , $\frac{5}{9}$ , $\frac{4}{7}$ . The fraction in the middle is $\frac{5}{9}$ .
A group of 24 students went on a field trip. One-third ( $\frac{1}{3}$ ) of the students chose to visit the dinosaur exhibit. Three-eighths ( $\frac{3}{8}$ ) of the students chose the space exhibit. Which exhibit was chosen by more students?
Answer: The space exhibit To solve this, we must compare the fractions $\frac{1}{3}$ and $\frac{3}{8}$ . The total number of students, 24, is extra information. To compare the fractions directly, we find a common denominator, which is 24. Dinosaur exhibit: $\frac{1}{3} = \frac{8}{24}$ . Space exhibit: $\frac{3}{8} = \frac{9}{24}$ . Since $9 > 8$ , $\frac{9}{24} > \frac{8}{24}$ , which means a larger fraction of students chose the space exhibit. Therefore, the space exhibit was chosen by more students.
Four friends are painting a long fence. After one hour, Liam has painted $\frac{3}{4}$ of his section, Noah has painted $\frac{2}{3}$ of his section, Olivia has painted $\frac{5}{6}$ of her section, and Emma has painted $\frac{7}{12}$ of her section. All sections are the same size. Who is in the lead, having painted the most?
Answer: Olivia To determine who is in the lead, we must find the largest fraction among $\frac{3}{4}$ , $\frac{2}{3}$ , $\frac{5}{6}$ , and $\frac{7}{12}$ . A common denominator for these fractions is 12. Convert each fraction: Liam: $\frac{3}{4} = \frac{9}{12}$ . Noah: $\frac{2}{3} = \frac{8}{12}$ . Olivia: $\frac{5}{6} = \frac{10}{12}$ . Emma: $\frac{7}{12}$ . Comparing the numerators, 10 is the largest. Therefore, Olivia has painted the most and is in the lead.
Which of the following expressions results in the largest value?
Answer: $\frac{1}{3} + \frac{1}{12}$ When comparing fractions through addition and subtraction, you need to find common denominators and calculate the actual values to determine which expression yields the largest result. Let's work through each expression systematically. For choice A: $\frac{1}{2} - \frac{1}{10}$ , convert to the common denominator 10: $\frac{5}{10} - \frac{1}{10} = \frac{4}{10} = 0.4$ For choice B: $\frac{1}{5} + \frac{1}{20}$ , convert to the common denominator 20: $\frac{4}{20} + \frac{1}{20} = \frac{5}{20} = 0.25$ For choice C: $\frac{3}{4} - \frac{1}{2}$ , convert to the common denominator 4: $\frac{3}{4} - \frac{2}{4} = \frac{1}{4} = 0.25$ For choice D: $\frac{1}{3} + \frac{1}{12}$ , convert to the common denominator 12: $\frac{4}{12} + \frac{1}{12} = \frac{5}{12} ≈ 0.417$ Comparing the results: A gives 0.4, B gives 0.25, C gives 0.25, and D gives approximately 0.417. Choice D produces the largest value. Choice A is close but falls short of D's value. Choices B and C both equal 0.25, making them tied for the smallest values. The key trap here is that addition doesn't automatically create larger results than subtraction—the actual fractional values matter more than the operations. When comparing fraction expressions, always calculate the final decimal values rather than making assumptions based on whether you're adding or subtracting. Convert everything to a common form for easy comparison.
If a positive whole number is added to both the numerator and the denominator of the fraction $\frac{2}{5}$ , how does the new fraction compare to $\frac{2}{5}$ ?
Answer: The new fraction is larger than $\frac{2}{5}$ . Let's test this concept by adding a positive whole number, for example, 1. The new fraction becomes $\frac{2+1}{5+1} = \frac{3}{6} = \frac{1}{2}$ . To compare $\frac{1}{2}$ with $\frac{2}{5}$ , we use a common denominator of 10. $\frac{1}{2} = \frac{5}{10}$ and $\frac{2}{5} = \frac{4}{10}$ . Since $\frac{5}{10} > \frac{4}{10}$ , the new fraction is larger. This pattern holds true for any positive proper fraction; adding the same positive number to the numerator and denominator increases its value, bringing it closer to 1.
David and Sarah went for a run. David ran $2\frac{3}{5}$ miles and Sarah ran $2\frac{2}{3}$ miles. Who ran a farther distance?
Answer: Sarah ran farther. To compare the mixed numbers $2\frac{3}{5}$ and $2\frac{2}{3}$ , notice that the whole number part (2) is the same for both. Therefore, we only need to compare the fractional parts, $\frac{3}{5}$ and $\frac{2}{3}$ . Find a common denominator, which is 15. Convert the fractions: $\frac{3}{5} = \frac{9}{15}$ and $\frac{2}{3} = \frac{10}{15}$ . Since $\frac{10}{15} > \frac{9}{15}$ , Sarah's fractional distance is greater. This means Sarah ran farther than David.
At a school field day, the fifth-grade class completed $\frac{5}{8}$ of the events before lunch. The sixth-grade class completed $\frac{2}{3}$ of the events. If both grades had the same number of events, which grade was closer to being finished?
Answer: The sixth-grade class When comparing fractions to see which represents being "closer to finished," you need to determine which fraction is larger. Since both classes had the same number of events, you can directly compare $\frac{5}{8}$ and $\frac{2}{3}$ . To compare fractions with different denominators, find a common denominator. The least common multiple of 8 and 3 is 24. Converting both fractions: $\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}$ and $\frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24}$ . Since $\frac{16}{24} > \frac{15}{24}$ , the sixth-grade class completed more of their events and was closer to being finished. Choice A is incorrect because $\frac{5}{8}$ is actually smaller than $\frac{2}{3}$ , so the fifth-grade class was further from completion. Choice B is wrong because the problem states both grades had the same number of events, making the total number irrelevant to the comparison. Choice C is incorrect because the fractions are not equal—when converted to the same denominator, $\frac{15}{24} \neq \frac{16}{24}$ . Choice D is correct because $\frac{2}{3} > \frac{5}{8}$ . When comparing fractions on the ISEE, always convert to a common denominator or use cross-multiplication to avoid errors. Don't let different denominators fool you into thinking you can't make a direct comparison—there's always a way to determine which fraction is larger.
A baker used $\frac{11}{4}$ cups of sugar for a large cake and $2\frac{1}{2}$ cups of sugar for a batch of cookies. Which dessert required more sugar?
Answer: The cake required more sugar. To compare $\frac{11}{4}$ and $2\frac{1}{2}$ , we should convert them to the same format. Let's convert the improper fraction $\frac{11}{4}$ to a mixed number. Divide 11 by 4, which is 2 with a remainder of 3. So, $\frac{11}{4} = 2\frac{3}{4}$ . Now we compare $2\frac{3}{4}$ (for the cake) with $2\frac{1}{2}$ (for the cookies). Since the whole numbers are both 2, we compare the fractions $\frac{3}{4}$ and $\frac{1}{2}$ . We know that $\frac{3}{4} > \frac{2}{4} = \frac{1}{2}$ . Therefore, the cake required more sugar.
Two identical pies are cut into slices. The first pie is cut into 10 equal slices, and the second pie is cut into 12 equal slices. If one person takes 3 slices from the first pie and another person takes 3 slices from the second pie, which statement is true?
Answer: The person who took slices from the pie cut into 10 slices got more pie. The first person takes $\frac{3}{10}$ of a pie. The second person takes $\frac{3}{12}$ of a pie. We need to compare $\frac{3}{10}$ and $\frac{3}{12}$ . When two fractions have the same numerator, the fraction with the smaller denominator is larger, because the whole is divided into fewer, larger pieces. Since $10 < 12$ , it follows that $\frac{3}{10} > \frac{3}{12}$ . Therefore, the person who took slices from the pie cut into 10 slices got more pie.
In a bag of marbles, $\frac{2}{7}$ are red and $\frac{1}{4}$ are blue. The rest of the marbles are green. Which statement correctly compares the number of red and blue marbles?
Answer: There are more red marbles than blue marbles. The question asks to compare the number of red and blue marbles. This requires comparing the fractions $\frac{2}{7}$ (red) and $\frac{1}{4}$ (blue). We can use cross-multiplication: for $\frac{2}{7}$ , the product is $2 \times 4 = 8$ . For $\frac{1}{4}$ , the product is $1 \times 7 = 7$ . Since $8 > 7$ , the fraction $\frac{2}{7}$ is greater than $\frac{1}{4}$ . Therefore, there are more red marbles than blue marbles. The information about green marbles is extra and not needed to answer the question asked.
Which of the following fractions is closest in value to $\frac{1}{2}$ ?
Answer: $\frac{5}{9}$ To find which fraction is closest to $\frac{1}{2}$ , we find the absolute difference between each fraction and $\frac{1}{2}$ . A) $|\frac{3}{8} - \frac{4}{8}| = \frac{1}{8}$ . B) $|\frac{4}{7} - \frac{1}{2}| = |\frac{8}{14} - \frac{7}{14}| = \frac{1}{14}$ . C) $|\frac{5}{9} - \frac{1}{2}| = |\frac{10}{18} - \frac{9}{18}| = \frac{1}{18}$ . D) $|\frac{7}{12} - \frac{1}{2}| = |\frac{7}{12} - \frac{6}{12}| = \frac{1}{12}$ . Now we must find the smallest of these differences: $\frac{1}{8}, \frac{1}{14}, \frac{1}{18}, \frac{1}{12}$ . When fractions have the same numerator (1 in this case), the one with the largest denominator is the smallest. The largest denominator is 18, so $\frac{1}{18}$ is the smallest difference. Therefore, $\frac{5}{9}$ is closest to $\frac{1}{2}$ .
A turtle takes $\frac{1}{5}$ of an hour to cross a yard. A snail takes $\frac{2}{9}$ of an hour to cross the same yard. Which animal is faster?
Answer: The turtle is faster. The faster animal is the one that takes less time to travel the same distance. We need to compare the times $\frac{1}{5}$ hour and $\frac{2}{9}$ hour and find the smaller value. Using cross-multiplication to compare the fractions: for $\frac{1}{5}$ , the product is $1 \times 9 = 9$ ; for $\frac{2}{9}$ , the product is $2 \times 5 = 10$ . Since $9 < 10$ , it means $\frac{1}{5} < \frac{2}{9}$ . The turtle takes less time, so the turtle is faster.
A recipe calls for an amount of sugar that is more than $\frac{1}{3}$ cup but less than $\frac{1}{2}$ cup. Which of the following amounts of sugar could be used?
Answer: $\frac{5}{12}$ cup To find a fraction between $\frac{1}{3}$ and $\frac{1}{2}$ , convert them to fractions with a common denominator. A common denominator for all the fractions is 60. $\frac{1}{3} = \frac{20}{60}$ and $\frac{1}{2} = \frac{30}{60}$ . We need a fraction between $\frac{20}{60}$ and $\frac{30}{60}$ . Let's convert the answer choices: A) $\frac{1}{4} = \frac{15}{60}$ (too small). B) $\frac{5}{12} = \frac{25}{60}$ (this is between $\frac{20}{60}$ and $\frac{30}{60}$ ). C) $\frac{2}{3} = \frac{40}{60}$ (too large). D) $\frac{3}{5} = \frac{36}{60}$ (too large). Therefore, $\frac{5}{12}$ is the correct amount.
A bookshelf is $\frac{7}{9}$ full. Another bookshelf of the same size is $\frac{3}{4}$ full. Which bookshelf has more empty space?
Answer: The bookshelf that is $\frac{3}{4}$ full. First, calculate the empty space for each bookshelf. The first bookshelf's empty space is $1 - \frac{7}{9} = \frac{2}{9}$ . The second bookshelf's empty space is $1 - \frac{3}{4} = \frac{1}{4}$ . Now, compare the fractions of empty space, $\frac{2}{9}$ and $\frac{1}{4}$ . Using a common denominator of 36: $\frac{2}{9} = \frac{8}{36}$ and $\frac{1}{4} = \frac{9}{36}$ . Since $\frac{9}{36} > \frac{8}{36}$ , the bookshelf that is $\frac{3}{4}$ full has more empty space.
On Monday, Chloe ate $\frac{1}{2}$ of a small pizza. On Tuesday, she ate $\frac{1}{3}$ of a large pizza. Which of the following must be true about the amount of pizza Chloe ate?
Answer: It cannot be determined who ate more pizza. The fractions refer to different-sized wholes (a 'small pizza' versus a 'large pizza'). Because the total sizes of the pizzas are not the same and are not specified, we cannot compare the absolute amounts of pizza eaten. For example, $\frac{1}{3}$ of a very large pizza could be more than $\frac{1}{2}$ of a very small pizza. Without more information about the sizes of the pizzas, no definitive comparison can be made.
In a school election for class president, Candidate A received $\frac{9}{10}$ of the votes in Mr. Smith's class. In Ms. Jones's class of the same size, Candidate B received $\frac{13}{15}$ of the votes. Which statement correctly compares the results?
Answer: Candidate A received a greater fraction of the votes. To compare $\frac{9}{10}$ and $\frac{13}{15}$ , we can find a common denominator, which is 30. Convert the fractions: Candidate A: $\frac{9}{10} = \frac{27}{30}$ . Candidate B: $\frac{13}{15} = \frac{26}{30}$ . Since $27 > 26$ , we know that $\frac{27}{30} > \frac{26}{30}$ . Therefore, Candidate A received a greater fraction of the votes.
Marco ate $\frac{2}{5}$ of his pizza. Jada ate $\frac{3}{8}$ of her pizza, which was the same size as Marco's. Who has more pizza left over?
Answer: Jada has more pizza left over. First, determine the fraction of pizza each person has left. Marco has $1 - \frac{2}{5} = \frac{3}{5}$ of his pizza left. Jada has $1 - \frac{3}{8} = \frac{5}{8}$ of her pizza left. Next, compare these two fractions. To compare $\frac{3}{5}$ and $\frac{5}{8}$ , find a common denominator, which is 40. Marco's remaining pizza is $\frac{3}{5} = \frac{24}{40}$ . Jada's remaining pizza is $\frac{5}{8} = \frac{25}{40}$ . Since $\frac{25}{40} > \frac{24}{40}$ , Jada has more pizza left over.

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